1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Eduardwww [97]
3 years ago
9

Consider the reaction

Chemistry
1 answer:
SOVA2 [1]3 years ago
5 0

Answer :

(a) The average rate will be:

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

(b) The average rate will be:

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

Thus, the rate of reaction will be:

\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}

<u>Part (a) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}

and,

\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s

\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s

<u>Part (b) :</u>

<u>Given:</u>

\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s

As,  

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

and,

-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s

\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s

You might be interested in
Complete the statement to describe the calcium concentration that must be maintained in human body cells. ( fill in the blank )T
erik [133]

Answer:

The calcium concentration must be greater outside the cell than inside the cell.

Explanation:

My previous answer was deleted from the explanation I provided from another website.

7 0
3 years ago
Read 2 more answers
molar mass of beryllium (Be) molar mass of silicon (Si) molar mass of calcium (Ca) molar mass of rhodium (Rh)
Neko [114]
Molarmass of beryllium is 9.0
molar mass of silicon is 28.4
molar mass of calcium is 40.1
molar mass of rhodium is 103.
3 0
3 years ago
Read 2 more answers
What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!
Lemur [1.5K]
The answer to your question is nitrogen dioxide

8 0
3 years ago
If two elements are from the
Vladimir79 [104]

Answer:

The element from Group 13.

Explanation:

Following trends of the periodic table, atomic radius of the elements increase going down from the right side and decrease on its way up diagonally to the left. (sry if u can't understand me)

7 0
3 years ago
Which of these compounds would most likely be found in a deposit of natural gas?
Valentin [98]
The correct answer for the given question above would be option A. The compound that would most likely be found in a deposit of natural gas is CH4 or METHANE. Methane is the main constituent of natural gas. It is<span> a colorless, odorless gas with a wide distribution in nature. Hope this answers your question.</span>
8 0
3 years ago
Read 2 more answers
Other questions:
  • To show the electron configuration for an atom, when would it be better to use an orbital notation than to use a written configu
    8·2 answers
  • . A standard dry cell has an output voltage of A. 1.5 VDC. B. 1.1 VDC. C. 1.2 VDC. D. 2.0 VDC.
    11·1 answer
  • Will make brainliest if answer correctly.
    7·1 answer
  • Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should
    9·1 answer
  • Ammonia and oxygen react to form nitrogen monoxide and water, like this:
    15·1 answer
  • What can you determine about an element if you only know its group number on the periodic table?
    6·2 answers
  • What is a hypothesis?
    10·1 answer
  • 1. Which of the following best describes natural selection?
    7·1 answer
  • Which energy changes would take place when a water is heated using a Bunsen burner?
    11·1 answer
  • Pls help I will give u points pls
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!